# 點算的奧秘：指數母函數

《點算的奧秘：普通母函數》中，筆者介紹了形如

Σ0 ≤ r < ∞ arxr

「指數母函數」的有用之處在於，我們可以使用與「普通母函數」完全對應的方法來解決「排列」問題。在介 紹「普通母函數」時，筆者曾指出我們可以把「普通母函數」中的xr看成代表從某類物件中抽取k 個出來。現在我們同樣可以把「指數母函數」中的xr / r!看成代表從某類物件中抽k個出來排列。 舉例說，(1 + x + x2/2! + x3/3!)(註1)便代表可從某類物件中抽0至3個(即最多3個) 出來排列；(x2/2! + x3/3! + ...)則代表可從某類物件中抽2個、3個以至無限個(即 最少2個)出來排列；而(x3/3! + x4/4!)則代表可從某類物件中抽3個或4個(即上限為4 個、下限為3個)出來排列。

(1 + x)n = (x + 1)n = Σ0 ≤ r ≤ nC(n, r) xr

(1 + x)n = Σ0 ≤ r ≤ nP(n, r) xr / r!

(1 + x + ... xr1/r1!) × (1 + x + ... xr2/r2!) × ... (1 + x + ... xrn/rn!)    (1)

 (xr1/r1!) × (xr2/r2!) × ... (xrn/rn!) = xr / (r1! × r2! × ... rn!)

r! / (r1! × r2! × ... rn!) × xr / r!

r! / (r1! × r2! × ... rn!)

(1 + x + x2/2! + x3/3!) × (1 + x + x2/2!) × (1 + x)

x4/2! + x4/2! + x4/(2!2!) + x4/3! + x4/3! = 19x4/12 = 38x4/4!

Σ0 ≤ r < ∞ xr / r! = 1 + x + x2/2! + x3/3! + ... = ex    (2)

(1 + x + x2/2! + x3/3! + ...)n = enx

 enx = Σ0 ≤ r < ∞ (nx)r / r! = Σ0 ≤ r < ∞ nr xr / r!

nr

Σ1 ≤ r < ∞ xr / r! = x + x2/2! + x3/3! + ... = ex − 1    (3)

Σ2 ≤ r < ∞ xr / r! = x2/2! + x3/3! + ... = ex − 1 − x    (4)

 (ex + e−x) / 2 = (1 + x + x2/2! + x3/3! + x4/4! + ... 1 − x + x2/2! − x3/3! + x4/4! + ...) / 2 = (2 + 2x2/2! + 2x4/4! + ...) / 2 = 1 + x2/2! + x4/4! + ... (ex − e−x) / 2 = (1 + x + x2/2! + x3/3! + x4/4! + ... − 1 + x − x2/2! + x3/3! − x4/4! + ...) / 2 = (2x + 2x3/3! + 2x5/5! + ...) / 2 = 1 + x3/3! + x5/5! + ...

Σ0 ≤ r < ∞ x2r / (2r)! = 1 + x2/2! + x4/4! + ... = (ex + e−x) / 2    (5)

Σ0 ≤ r < ∞ x2r+1 / (2r+1)! = x + x3/3! + x5/5! + ... = (ex − e−x) / 2    (6)

 ex × (ex − e−x) / 2 × (ex + e−x) / 2 = (e3x − e−x) / 4 = [Σ0 ≤ r < ∞ (3x)r / r! − Σ0 ≤ r < ∞ (−x)r / r!] / 4 = Σ0 ≤ r < ∞ [(3r − (−1)r) / 4] xr / r!

 (1 + x + x2/2! + x3/3!) × (ex − 1) × ex = (1 + x + x2/2! + x3/3!) × (e2x − ex) = (1 + x + x2/2! + x3/3!) × [Σ0 ≤ r < ∞ (2x)r / r! − Σ0 ≤ r < ∞ xr / r!] = (1 + x + x2/2! + x3/3!) × Σ0 ≤ r < ∞ (2r − 1) xr / r!

 1 × (24 − 1) x4/4! + x × (23 − 1) x3/3! + x2/2! × (22 − 1) x2/2! + x3/3! × (21 − 1) x = 15x4/4! + 7x4/3! + 3x4/(2!2!) + x4/3! = 15x4/4! + 28x4/4! + 18x4/4! + 4x4/4! = 65x4/4!

 (x + x2/2! + ...)n = (ex − 1)n (應用公式(3)) = Σ0 ≤ k ≤ n C(n, k) (−1)k ex(n−k) (應用「二項式定理」) = Σ0 ≤ k ≤ n C(n, k) (−1)k Σ0 ≤ r < ∞ (n − k)r xr / r! (應用公 式(2)) = Σ0 ≤ r < ∞ [Σ0 ≤ k ≤ n (−1)k C(n, k) (n − k)r] xr / r!

Σ0 ≤ k ≤ n (−1)k C(n, k) (n − k)r    (7)

Σ0 ≤ r < ∞ x2r = 1 + x2 + x4 + ... = 1 / (1 − x2)

Σ0 ≤ r < ∞ x2r+1 = x + x3 + x5 + ... = x / (1 − x2)

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